Let $g$ be a twice differentiable function, and let $g(-1)=4$, $g'(-1)=0$, and $g''(-1)=3$. What occurs in the graph of $g$ at the point $(-1,4)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-1,4)$ is a minimum point. (Choice B) B $(-1,4)$ is a maximum point. (Choice C) C There's not enough information to tell.
Answer: Since $g'(-1)=0$, we know that $x=-1$ is a critical point. The second derivative test allows us to analyze what happens in the graph of $g$ at this point according to these three cases: If $g''(-1)>0$, the graph of $g$ has a minimum point at $x=-1$. If $g''(-1)<0$, the graph of $g$ has a maximum point at $x=-1$. If $g''(-1)=0$, the test is inconclusive. [Why is this so?] We are given that $g''(-1)=3>0$. Therefore, $(-1,4)$ is a minimum point.